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Proof step 3 from proof lookup-01 (previous step)

∀ a, d. (ax0 + (a × d)')[a] = d ∨ a ∈ ax0 ⊦ ((p + ax0) + (a × d)')[a] = d ∨ a ∈ p + ax0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ a, d. (ax0 + (a × d)')[a] = d ∨ a ∈ ax0 ⊦ ((p ' + ax0) + (a × d)')[a] = d, a ∈ ax0, a ∈ p '

Rule argument

Additional information

Used simplifier rules: inapp: ⊦ a ∈ ax + ay ↔ ¬ ¬(a ∈ ax ∨ a ∈ ay);% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;% ⊦ p + ax = p ' + ax;